Let $C_{1}, C_{2}, C_{3}, ......$ be the circles with radius $r_{1}$ such that equation of circle $C_{1} : x^{2} + y^{2} = t_{1}^{2} .$ Circle $C_{1}$ intersects positive x-axis and y-axis at $A_{1}$ and $B_{1}$ respectively. There exist a point $P_{i},$ perpendicular from $P_{i}$ on $O A_{i}$ and $O B_{i}$ are points $D_{i}$ and $E_{i}$ respectively such that $O D_{i} : O E_{i} = 3 : 4$ and there exist a point $Q_{i}$ on circle $C_{i + 1,}$ if $A_{i} Q_{i} = P_{i} Q_{i}$ and $∠ A_{i} Q_{i} P_{i} = \frac{π}{2} .$ If $S_{i}$ be the area of triangle $A_{i} Q_{i} P_{i}$ and $r_{1} > r_{2} > r_{3} > ..... > r_{n,}$ where $r_{1} = 1,$ then $\lim_{n \to \infty} (\sum_{i = 1}^{n} S_{i})$ is less than or equal to :

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$\frac{1}{8}$

$\frac{5}{8}$

$\frac{3}{4}$

$\frac{1}{4}$

answer is A, B, D.

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Detailed Solution

$A_{i} P_{i} = \frac{2}{\sqrt{5}} r_{i} a n d r_{i + 1} = \frac{r_{i}}{\sqrt{5}}$
And $S_{i} = \frac{r_{1}^{2}}{5} \Rightarrow S_{i + 1} = \frac{r_{i + 1}^{2}}{5} = \frac{r_{1}^{2}}{25}$
$\Rightarrow \frac{S_{i + 1}}{S_{i}} = \frac{1}{5}$
$\lim_{n \to \infty} (S_{1} + S_{2} + S_{3} + ..... + S_{n}) = \lim_{n \to \infty} (\frac{\frac{1}{5}}{1 - \frac{1}{5}}) = \frac{1}{4}$